## 物理代写|量子场论代写Quantum field theory代考|PHYC90008

statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。

## 物理代写|量子场论代写Quantum field theory代考|Classical electrodynamics

We would expect the Hamiltonian of a system of moving charges, such as an atom, in an electromagnetic field to consist of three parts: a part referring to matter (i.e. the charges), a part referring to the electromagnetic field and a part describing the interaction between matter and field.

For a system of point masses $m_i, i=1, \ldots, N$, with charges $e_i$ and position coordinates $\mathbf{r}i$, the Hamiltonian is $$H{\mathrm{m}}=\sum_i \frac{\mathbf{p}i^2}{2 m_i}+H{\mathrm{C}}$$
where $H_{\mathrm{C}}$ is the Coulomb interaction
$$H_{\mathrm{C}} \equiv \frac{1}{2} \sum_{\substack{i, j \(i \neq j)}} \frac{e_i e_j}{4 \pi\left|\mathbf{r}_i-\mathbf{r}_j\right|}$$
and $\mathbf{p}_i=m_i \mathrm{~d} \mathbf{r}_i / \mathrm{d} t$ is the kinetic momentum of the $i$ th particle. This is the usual Hamiltonian of atomic physics, for example.

The electromagnetic field in interaction with charges is described by Maxwell’s equations [Eqs. (1.1)]. We continue to use the Coulomb gauge, $\boldsymbol{\nabla} \cdot \mathbf{A}=0$, so that the electric field (1.2) decomposes into transverse and longitudinal fields
$$\mathbf{E}=\mathbf{E}{\mathbf{T}}+\mathbf{E}{\mathbf{L}},$$
where
$$\mathbf{E}{\mathbf{T}}=-\frac{1}{c} \frac{\partial \mathbf{A}}{\partial t}, \quad \mathbf{E}{\mathbf{L}}=-\nabla \phi$$

## 物理代写|量子场论代写Quantum field theory代考|Quantum electrodynamics

The quantization of the system described by the Hamiltonian (1.63) is carried out by subjecting the particles’ coordinates $\mathbf{r}i$ and canonically conjugate momenta $\mathbf{p}_i$ to the usual commutation relations (e.g. in the coordinate representation $\mathbf{p}_i \rightarrow i \hbar \boldsymbol{\nabla}_i$ ), and quantizing the radiation field, as in Section 1.2.3. The longitudinal electric field $\mathbf{E}{\mathbf{L}}$ does not pro

The eigenstates of $H_0$ are again of the form
$$\left|A, \ldots n_r(\mathbf{k}) \ldots\right\rangle=|A\rangle\left|\ldots n_r(\mathbf{k}) \ldots\right\rangle,$$
with $|A\rangle$ and $\left|\ldots n_r(\mathbf{k}) \ldots\right\rangle$ eigenstates of $H_{\mathrm{m}}$ and $H_{\mathrm{rad}}$.
Compared with the electric dipole interaction (1.40), the interaction (1.62) differs in that it contains a term quadratic in the vector potential. This results in two-photon processes in first-order perturbation theory (i.e. emission or absorption of two photons or scattering). In addition, the first term in (1.62) contains magnetic interactions and higher-order effects due to the spatial variation of $\mathbf{A}(\mathbf{x}, t)$, which are absent from the electric dipole interaction (1.40). These aspects are illustrated in the applications to radiative transitions and Thomson scattering which follow.

vide any additional degrees of freedom, being completely determined via the first Maxwell equation $\boldsymbol{\nabla} \cdot \mathbf{E}_{\mathbf{L}}=\rho$ by the charges.

The interaction $H_{\mathrm{I}}$ in Eq, (1.63) is usually treated as a perturbation which causes transitions between the states of the non-interacting Hamiltonian
$$H_0=H_{\mathrm{m}}+H_{\text {rad }} \text {. }$$

## 物理代写|量子场论代写Quantum field theory代考|Classical electrodynamics

$$H_{\mathrm{C}} \equiv \frac{1}{2} \sum_{\substack{i, j (i \neq j)}} \frac{e_i e_j}{4 \pi\left|\mathbf{r}_i-\mathbf{r}_j\right|}$$
$\mathbf{p}_i=m_i \mathrm{~d} \mathbf{r}_i / \mathrm{d} t$是第$i$个粒子的动能。例如，这是原子物理学中常用的哈密顿量。

$$\mathbf{E}=\mathbf{E}{\mathbf{T}}+\mathbf{E}{\mathbf{L}},$$

$$\mathbf{E}{\mathbf{T}}=-\frac{1}{c} \frac{\partial \mathbf{A}}{\partial t}, \quad \mathbf{E}{\mathbf{L}}=-\nabla \phi$$

## 物理代写|量子场论代写Quantum field theory代考|Quantum electrodynamics

$H_0$的特征态也是这种形式
$$\left|A, \ldots n_r(\mathbf{k}) \ldots\right\rangle=|A\rangle\left|\ldots n_r(\mathbf{k}) \ldots\right\rangle,$$

Eq，(1.63)中的相互作用$H_{\mathrm{I}}$通常被视为引起非相互作用哈密顿量状态之间转换的扰动
$$H_0=H_{\mathrm{m}}+H_{\text {rad }} \text {. }$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|量子场论代写Quantum field theory代考|PHYS4125

statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。

## 物理代写|量子场论代写Quantum field theory代考|The quantized radiation field

The harmonic oscillator results we have derived can at once be applied to the radiation field. Its Hamiltonian, Eq. (1.18), is a superposition of independent harmonic oscillator Hamiltonians (1.20), one for each mode of the radiation field. [The order of the factors in (1.18) is not significant and can be changed, since the $a_{\mathrm{r}}$ and $a_r^*$ are classical amplitudes.] We therefore introduce commutation relations analogous to Eq. (1.19)
$$\left.\begin{array}{l} {\left[a_r(\mathbf{k}), a_s^{\dagger}\left(\mathbf{k}^{\prime}\right)\right]=\delta_{r s} \delta_{\mathbf{k} \mathbf{k}^{\prime}}} \ {\left[a_r(\mathbf{k}), a_s\left(\mathbf{k}^{\prime}\right)\right]=\left[a_r^{\dagger}(\mathbf{k}), a_s^{\dagger}\left(\mathbf{k}^{\prime}\right)\right]=0} \end{array}\right}$$
and write the Hamiltonian (1.18) as
$$H_{\mathrm{rad}}=\sum_{\mathbf{k}} \sum_r \hbar \omega_{\mathbf{k}}\left(a_r^{\dagger}(\mathbf{k}) a_r(\mathbf{k})+\frac{1}{2}\right) .$$
The operators
$$N_r(\mathbf{k})=a_r^{\dagger}(\mathbf{k}) a_r(\mathbf{k})$$
then have eigenvalues $n_r(\mathbf{k})=0,1,2, \ldots$, and eigenfunctions of the form (1.25)
$$\left|n_r(\mathbf{k})\right\rangle=\frac{\left[a_r^{\dagger}(\mathbf{k})\right]^{n_r(\mathbf{k})}}{\sqrt{n_r(\mathbf{k}) !}}|0\rangle .$$
The eigenfunctions of the radiation Hamiltonian (1.30) are products of such states, i.e.
$$\left|\ldots n_r(\mathbf{k}) \ldots\right\rangle=\prod_{\mathbf{k}i} \prod{r_i}\left|n_{r_i}\left(\mathbf{k}i\right)\right\rangle$$ with energy $$\sum{\mathbf{k}} \sum_r \hbar \omega_{\mathbf{k}}\left(n_r(\mathbf{k})+\frac{1}{2}\right)$$

## 物理代写|量子场论代写Quantum field theory代考|The Electric Dipole Interaction

In the last section we quantized the radiation field. Since the occupation number operators $a_r^{\dagger}(\mathbf{k}) a_r(\mathbf{k})$ commute with the radiation Hamiltonian (1.37), the occupation numbers $n_r(\mathbf{k})$ are constants of the motion for the free field. For anything ‘to happen’ requires interactions with charges and currents so that photons can be absorbed, emitted or scattered.

The complete description of the interaction of a system of charges (for example, an atom or a nucleus) with an electromagnetic field is very complicated. In this section we shall consider the simpler and, in practice, important special case of the interaction occurring via the electric dipole moment of the system of charges. The more complete (but still noncovariant) treatment of Section 1.4 will justify some of the points asserted in this section.
We shall consider a system of $N$ charges $e_1, e_2, \ldots, e_N$ which can be described nonrelativistically, i.e. the position of $\mathrm{e}i, i=1, \ldots, N$, at time $t$ is classically given by $\mathbf{r}_i=\mathbf{r}_i(t)$. We consider transitions between def inite initial and final states of the system (e.g. between two states of an atom). The transitions are brought about by the electric dipole interaction if two approximations are valid. Firstly it is permissible to neglect the interactions with the magnetic field. Secondly, one may neglect the spatial variation of the electric radiation field, causing the transitions, across the system of charges (e.g. across the atom). Under these conditions the electric field $$\mathbf{E}{\mathrm{T}}(\mathbf{r}, t)=-\frac{1}{c} \frac{\partial \mathbf{A}(\mathbf{r}, t)}{\partial t},$$
resulting from the transverse vector potential (1.38) of the radiation field (we are again using the Coulomb gauge $\boldsymbol{\nabla} \cdot \mathbf{A}=0$ ), can be calculated at one point somewhere inside the system of charges, instead of at the position of each charge. ${ }^6$ Taking this point as the origin of coordinates $\mathbf{r}=0$, we obtain for the interaction causing transitions, the electric dipole interaction $H_{\mathrm{I}}$ given by
$$H_{\mathrm{I}}=-\mathbf{D} \cdot \mathbf{E}_{\mathrm{T}}(0, t)$$
where the electric dipole moment is defined by
$$\mathbf{D}=\sum e_i \mathbf{r}_i$$

## 物理代写|量子场论代写Quantum field theory代考|The quantized radiation field

$$\left.\begin{array}{l} {\left[a_r(\mathbf{k}), a_s^{\dagger}\left(\mathbf{k}^{\prime}\right)\right]=\delta_{r s} \delta_{\mathbf{k} \mathbf{k}^{\prime}}} \ {\left[a_r(\mathbf{k}), a_s\left(\mathbf{k}^{\prime}\right)\right]=\left[a_r^{\dagger}(\mathbf{k}), a_s^{\dagger}\left(\mathbf{k}^{\prime}\right)\right]=0} \end{array}\right}$$

$$H_{\mathrm{rad}}=\sum_{\mathbf{k}} \sum_r \hbar \omega_{\mathbf{k}}\left(a_r^{\dagger}(\mathbf{k}) a_r(\mathbf{k})+\frac{1}{2}\right) .$$

$$N_r(\mathbf{k})=a_r^{\dagger}(\mathbf{k}) a_r(\mathbf{k})$$

$$\left|n_r(\mathbf{k})\right\rangle=\frac{\left[a_r^{\dagger}(\mathbf{k})\right]^{n_r(\mathbf{k})}}{\sqrt{n_r(\mathbf{k}) !}}|0\rangle .$$

$$\left|\ldots n_r(\mathbf{k}) \ldots\right\rangle=\prod_{\mathbf{k}i} \prod{r_i}\left|n_{r_i}\left(\mathbf{k}i\right)\right\rangle$$ with energy $$\sum{\mathbf{k}} \sum_r \hbar \omega_{\mathbf{k}}\left(n_r(\mathbf{k})+\frac{1}{2}\right)$$

## 物理代写|量子场论代写Quantum field theory代考|The Electric Dipole Interaction

$$H_{\mathrm{I}}=-\mathbf{D} \cdot \mathbf{E}_{\mathrm{T}}(0, t)$$

$$\mathbf{D}=\sum e_i \mathbf{r}_i$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|量子场论代写Quantum field theory代考|FYS4170

statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。

## 物理代写|量子场论代写Quantum field theory代考|The classical field

Classical electromagnetic theory is summed up in Maxwell’s equations. In the presence of a charge density $\rho(\mathbf{x}, t)$ and a current density $\mathbf{j}(\mathbf{x}, \mathrm{t})$, the electric and magnetic fields $\mathbf{E}$ and $\mathbf{B}$ satisfy the equations
\begin{aligned} \boldsymbol{\nabla} \cdot \mathbf{E} & =\rho \ \boldsymbol{\nabla} \wedge \mathbf{B} & =\frac{1}{c} \mathbf{j}+\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} \ \boldsymbol{\nabla} \cdot \mathbf{B} & =0 \ \boldsymbol{\nabla} \wedge \mathbf{E} & =-\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t} \end{aligned}
where, as throughout this book, rationalized Gaussian (c.g.s.) units are being used. ${ }^1$
From the second pair of Maxwell’s equations [Eqs. (1.1c) and (1.1d)] follows the existence of scalar and vector potentials $\phi(\mathbf{x}, t)$ and $\mathbf{A}(\mathbf{x}, t)$, defined by
$$\mathbf{B}=\boldsymbol{\nabla} \wedge \mathbf{A}, \quad \mathbf{E}=-\boldsymbol{\nabla} \phi-\frac{1}{c} \frac{\partial \mathbf{A}}{\partial t} .$$
Eqs. (1.2) do not determine the potentials uniquely, since for an arbitrary function $f(\mathbf{x}, t)$ the transformation
$$\phi \rightarrow \phi^{\prime}=\phi+\frac{1}{c} \frac{\partial f}{\partial t}, \quad \mathbf{A} \rightarrow \mathbf{A}^{\prime}=\mathbf{A}-\nabla f$$
leaves the fields $\mathbf{E}$ and $\mathbf{B}$ unaltered. The transformation (1.3) is known as a gauge transformation of the second kind. Since all observable quantities can be expressed in terms of $\mathbf{E}$ and $\mathbf{B}$, it is a fundamental requirement of any theory formulated in terms of potentials that it is gauge-invariant, i.e. that the predictions for observable quantities are invariant under such gauge transformations.

Expressed in terms of the potentials, the second pair of Maxwell’s equations [Eqs. (1.1c) and (1.1d)] are satisfied automatically, while the first pair [Eqs. (1.1a) and (1.1b)] become
$$\begin{gathered} -\nabla^2 \phi-\frac{1}{c} \frac{\partial}{\partial t}(\boldsymbol{\nabla} \cdot \mathbf{A})=\square \phi-\frac{1}{c} \frac{\partial}{\partial t}\left(\frac{1}{c} \frac{\partial \phi}{\partial t}+\nabla \cdot \mathbf{A}\right)=\rho \ \square \mathbf{A}+\boldsymbol{\nabla}\left(\frac{1}{c} \frac{\partial \phi}{\partial t}+\nabla \cdot \mathbf{A}\right)=\frac{1}{c} \mathbf{j} \end{gathered}$$
where
$$\square \equiv \frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\nabla^2$$

## 物理代写|量子场论代写Quantum field theory代考|Harmonic oscillator

The harmonic oscillator Hamiltonian is, in an obvious notation,
$$H_{\mathrm{osc}}=\frac{p^2}{2 m}+\frac{1}{2} m \omega^2 q^2,$$
with $q$ and $p$ satisf ying the commutation relation $[q, p]=\mathrm{i} \hbar$. We introduce the operators
$$\left.\begin{array}{c} a \ a^{\dagger} \end{array}\right}=\frac{1}{(2 \hbar m \omega)^{1 / 2}}(m \omega q \pm \mathrm{i} p) .$$
These satisfy the commutation relation
$$\left[a, a^{\dagger}\right]=1,$$
and the Hamiltonian expressed in terms of $a$ and $a^{\dagger}$ becomes:
$$H_{\mathrm{osc}}=\frac{1}{2} \hbar \omega\left(a^{\dagger} a+a a^{\dagger}\right)=\hbar \omega\left(a^{\dagger} a+\frac{1}{2}\right) .$$
This is essentially the operator
$$N \equiv a^{\dagger} a,$$

which is positive definite, i.e. for any state $|\Psi\rangle$
$$\langle\Psi|N| \Psi\rangle=\left\langle\Psi\left|a^{\dagger} a\right| \Psi\right\rangle=\langle a \Psi \mid a \Psi\rangle \geq 0 .$$
Hence, $N$ possesses a lowest non-negative eigenvalue
$$\alpha_0 \geq 0 \text {. }$$
It follows from the eigenvalue equation
$$N|\alpha\rangle=\alpha|\alpha\rangle$$
and Eq. (1.19) that
$$N a|\alpha\rangle=(\alpha-1) a|\alpha\rangle, \quad N a^{\dagger}|\alpha\rangle=(\alpha+1) a^{\dagger}|\alpha\rangle,$$
i.e. $a|\alpha\rangle$ and $a^{\dagger}|\alpha\rangle$ are eigenfunctions of $N$ belonging to the eigenvalues $(\alpha-1)$ and $(\alpha+1)$, respectively. Since $\alpha_0$ is the lowest eigenvalue we must have
$$a\left|\alpha_0\right\rangle=0$$
and since
$$a^{\dagger} a\left|\alpha_0\right\rangle=\alpha_0\left|\alpha_0\right\rangle$$
Eq. (1.23) implies $\alpha_0=0$. It follows from Eqs. (1.19) and (1.22) that the eigenvalues of $N$ are the integers $n=0,1,2, \ldots$, and that if $\langle n \mid n\rangle=1$, then the states $|n \pm 1\rangle$, defined by
$$a|n\rangle=n^{1 / 2}|n-1\rangle, \quad a^{\dagger}|n\rangle=(n+1)^{1 / 2}|n+1\rangle,$$
are also normed to unity. If $\langle 0 \mid 0\rangle=1$, the normed eigenf unctions of $N$ are
$$|n\rangle=\frac{\left(a^{\dagger}\right)^n}{\sqrt{n !}}|0\rangle, \quad n=0,1,2, \ldots$$

## 物理代写|量子场论代写Quantum field theory代考|The classical field

\begin{aligned} \boldsymbol{\nabla} \cdot \mathbf{E} & =\rho \ \boldsymbol{\nabla} \wedge \mathbf{B} & =\frac{1}{c} \mathbf{j}+\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} \ \boldsymbol{\nabla} \cdot \mathbf{B} & =0 \ \boldsymbol{\nabla} \wedge \mathbf{E} & =-\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t} \end{aligned}

$$\mathbf{B}=\boldsymbol{\nabla} \wedge \mathbf{A}, \quad \mathbf{E}=-\boldsymbol{\nabla} \phi-\frac{1}{c} \frac{\partial \mathbf{A}}{\partial t} .$$

$$\phi \rightarrow \phi^{\prime}=\phi+\frac{1}{c} \frac{\partial f}{\partial t}, \quad \mathbf{A} \rightarrow \mathbf{A}^{\prime}=\mathbf{A}-\nabla f$$

$$\begin{gathered} -\nabla^2 \phi-\frac{1}{c} \frac{\partial}{\partial t}(\boldsymbol{\nabla} \cdot \mathbf{A})=\square \phi-\frac{1}{c} \frac{\partial}{\partial t}\left(\frac{1}{c} \frac{\partial \phi}{\partial t}+\nabla \cdot \mathbf{A}\right)=\rho \ \square \mathbf{A}+\boldsymbol{\nabla}\left(\frac{1}{c} \frac{\partial \phi}{\partial t}+\nabla \cdot \mathbf{A}\right)=\frac{1}{c} \mathbf{j} \end{gathered}$$

$$\square \equiv \frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\nabla^2$$

## 物理代写|量子场论代写Quantum field theory代考|Harmonic oscillator

$$H_{\mathrm{osc}}=\frac{p^2}{2 m}+\frac{1}{2} m \omega^2 q^2,$$

$$\left.\begin{array}{c} a \ a^{\dagger} \end{array}\right}=\frac{1}{(2 \hbar m \omega)^{1 / 2}}(m \omega q \pm \mathrm{i} p) .$$

$$\left[a, a^{\dagger}\right]=1,$$

$$H_{\mathrm{osc}}=\frac{1}{2} \hbar \omega\left(a^{\dagger} a+a a^{\dagger}\right)=\hbar \omega\left(a^{\dagger} a+\frac{1}{2}\right) .$$

$$N \equiv a^{\dagger} a,$$

$$\langle\Psi|N| \Psi\rangle=\left\langle\Psi\left|a^{\dagger} a\right| \Psi\right\rangle=\langle a \Psi \mid a \Psi\rangle \geq 0 .$$

$$\alpha_0 \geq 0 \text {. }$$

$$N|\alpha\rangle=\alpha|\alpha\rangle$$

$$N a|\alpha\rangle=(\alpha-1) a|\alpha\rangle, \quad N a^{\dagger}|\alpha\rangle=(\alpha+1) a^{\dagger}|\alpha\rangle,$$

$$a\left|\alpha_0\right\rangle=0$$

$$a^{\dagger} a\left|\alpha_0\right\rangle=\alpha_0\left|\alpha_0\right\rangle$$
Eq.(1.23)表示$\alpha_0=0$。由等式可知。(1.19)和式(1.22)可知$N$的特征值是整数$n=0,1,2, \ldots$，如果$\langle n \mid n\rangle=1$，则状态$|n \pm 1\rangle$，定义为
$$a|n\rangle=n^{1 / 2}|n-1\rangle, \quad a^{\dagger}|n\rangle=(n+1)^{1 / 2}|n+1\rangle,$$

$$|n\rangle=\frac{\left(a^{\dagger}\right)^n}{\sqrt{n !}}|0\rangle, \quad n=0,1,2, \ldots$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|量子场论代写Quantum field theory代考|SI2410

statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。

## 物理代写|量子场论代写Quantum field theory代考|Time reversal

Finally, let us turn to the most confusing of the discrete symmetries, time reversal. As a Lorentz transformation,
$$T: \quad(t, \vec{x}) \rightarrow(-t, \vec{x}) .$$
We are going to need a transformation of our spinor fields, $\psi$, such that (at least) the kinetic Lagrangian is invariant. To do this, we need $\bar{\psi} \gamma^\mu \psi$ to transform as a 4-vector under $T$, so that $i \bar{\psi} \not \partial \psi(t, \vec{x}) \rightarrow i \bar{\psi} \not \partial \psi(-t, x)$ and the action will be invariant. In particular, we need the 0 -component, $\bar{\psi} \gamma^0 \psi \rightarrow-\bar{\psi} \gamma^0 \psi$, which implies $\psi^{\dagger} \psi \rightarrow-\psi^{\dagger} \psi$. But this last form of the requirement is very odd – it says we need to turn a positive definite quantity into a negative definite quantity. This is impossible for any linear transformation $\psi \rightarrow \Gamma \psi$. Thus, we need to think harder.

We will discuss two possibilities. One we will call “simple $\hat{T}$,” and denote $\hat{T}$. It is the obvious parallel to parity. The other is the $T$ symmetry, which is normally what is meant by $T$ in the literature. This second $T$ was invented by Wigner in 1932 and requires $T$ to take $i \rightarrow-i$ in the whole Lagrangian in addition to acting on fields. While the simple $\hat{T}$ is the more natural generalization of the action of $T$ on 4-vectors, it is also kind of trivial. Wigner’s $T$ has important physical implications.

Before doing anything drastic, the simplest thing besides $T: \psi \rightarrow \Gamma \psi$ would be $T: \psi \rightarrow$ $\Gamma \psi^{\star}$, as with charge conjugation. We will call this transformation $\hat{T}$ to distinguish it from what is conventionally called $T$ in the literature. So,
$$\hat{T}: \quad \psi \rightarrow \Gamma \psi^{\star}, \quad \psi^{\dagger} \rightarrow\left(\Gamma \psi^{\star}\right)^{\dagger}=\psi^T \Gamma^{\dagger} .$$
That $\hat{T}$ should take particles to antiparticles is also understandable from the picture of antiparticles as particles moving backwards in time.

Then,
$$\psi^{\dagger} \psi \rightarrow \psi^T \Gamma^{\dagger} \Gamma \psi^{\star}=\Gamma_{\alpha \beta}^{\dagger} \Gamma_{\beta \gamma} \psi_\alpha \psi_\gamma^{\star}=-\psi_\gamma^{\star}\left(\Gamma_{\alpha \beta}^{\dagger} \Gamma_{\beta \gamma}\right)^T \psi_\alpha=-\psi^{\dagger}\left(\Gamma^{\dagger} \Gamma\right)^T \psi,$$
so we need $\Gamma^{\dagger} \Gamma=\mathbb{1}$, which says that $\Gamma$ is a unitary matrix. That is fine. But we also need $\bar{\psi} \gamma_i \psi$ and the mass term $\bar{\psi} \psi$ to be preserved. For the mass term,
$$\bar{\psi} \psi \rightarrow \psi^T \Gamma^{\dagger} \gamma_0 \Gamma \psi^{\star}=-\bar{\psi}\left(\Gamma^{\dagger} \gamma_0 \Gamma \gamma_0\right)^T \psi$$
This equals $\bar{\psi} \psi$ only if $\left{\Gamma, \gamma_0\right}=0$. Next,
$$\bar{\psi} \gamma_i \psi \rightarrow \psi^T \Gamma^{\dagger} \gamma_0 \gamma_i \Gamma \psi^{\star}=-\bar{\psi}\left(\Gamma^{\dagger} \gamma_0 \gamma_i \Gamma \gamma_0\right)^T \psi=-\bar{\psi}\left(\Gamma^{\dagger} \gamma_i \Gamma\right)^T \psi,$$
which should be equal to $\bar{\psi} \gamma_i \psi$ for $i=1,2,3$. So $\gamma_i \Gamma+\Gamma \gamma_i^T=0$, which implies $\left[\Gamma, \gamma_1\right]=$ $0,\left[\Gamma, \gamma_3\right]=0$ and $\left{\Gamma, \gamma_2\right}=0$. The unique (up to a constant) matrix that commutes with $\gamma_1$ and $\gamma_3$ and anticommutes with $\gamma_2$ and $\gamma_0$ is $\Gamma=\gamma_0 \gamma_2$. Thus,
$$\psi(t, \vec{x}) \rightarrow \gamma_0 \gamma_2 \psi^{\star}(-t, \vec{x}), \quad \psi^{\dagger}(t, \vec{x}) \rightarrow-\psi^T \gamma_2 \gamma_0(-t, \vec{x}) .$$

## 物理代写|量子场论代写Quantum field theory代考|Wigner’s T (i.e. what is normally called T )

What is normally called time reversal is a symmetry $T$ that was described in a 1932 paper by Wigner, and shown to be an explanation of Kramer’s degeneracy. To understand Kramer’s degeneracy, consider the Schrödinger equation,
$$i \partial_t \psi(t, \vec{x})=H \psi(t, \vec{x}),$$

where, for simplicity, let us say $H=\frac{p^2}{2 m}+V(x)$, which is real and time independent. If we take the complex conjugate of this equation and also $t \rightarrow-t$, we find
$$i \partial_t \psi^{\star}(-t, \bar{x})=H \psi^{\star}(-t, \vec{x}) .$$
Thus, $\psi^{\prime}(t, x)=\psi^{\star}(-t, x)$ is another solution to the Schrödinger equation. If $\psi$ is an energy eigenstate, then as long as $\psi \neq \xi \psi^{\star}$ for any complex number $\xi, \psi^{\prime}$ will be another state with the same energy. This doubling of states at each energy is known as Kramer’s degeneracy. In particular, for the hydrogen atom, $\psi_{n l m}(\vec{x})=R_n(r) Y_{l m}(\theta, \phi)$ are the energy eigenstates, so Kramer’s degeneracy says that the states with $m$ and $-m$ will be degenerate (which they are). The importance of this theorem is that it also holds for more complicated systems, and for systems in external electric fields, for which the exact eigenstates may not be known.

As we will soon see, this mapping, $\psi(t, \vec{x}) \rightarrow \psi^{\star}(-t, \vec{x})$, sends particles to particles (not antiparticles), unlike the simple $\hat{T}$ operator above. It has a nice interpretation: Suppose you made a movie of some physics process, then watched the movie backwards; time reversal implies you should not be able to tell which was “play” and which was “reverse.”

The trick to Wigner’s $T$ is that we had to complex conjugate and then take $\psi^{\prime}=\psi^{\star}$. This means in particular that the $i$ in the Schrödinger equation goes to $-i$ as well as the field transforming. This is the key to finding a way out of the problem that $\psi^{\dagger} \psi$ needed to flip sign under $T$, which we discussed at the beginning of the section. The kinetic term for $\psi$ is $i \bar{\psi} \gamma^0 \partial_0 \psi$; so if $i \rightarrow-i$ then, since $\partial_0 \rightarrow-\partial_0, \psi^{\dagger} \psi$ can be invariant. Thus we need
$$T: \quad i \rightarrow-i .$$

## 物理代写|量子场论代写Quantum field theory代考|Time reversal

$$T: \quad(t, \vec{x}) \rightarrow(-t, \vec{x}) .$$

$$\psi(t, \vec{x}) \rightarrow \gamma_0 \gamma_2 \psi^{\star}(-t, \vec{x}), \quad \psi^{\dagger}(t, \vec{x}) \rightarrow-\psi^T \gamma_2 \gamma_0(-t, \vec{x}) .$$

## 物理代写|量子场论代写Quantum field theory代考|Wigner’s T (i.e. what is normally called T )

$$i \partial_t \psi(t, \vec{x})=H \psi(t, \vec{x}),$$

$$i \partial_t \psi^{\star}(-t, \bar{x})=H \psi^{\star}(-t, \vec{x}) .$$

$$T: \quad i \rightarrow-i .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 物理代写|量子场论代写Quantum field theory代考|PHYS4260

statistics-lab™ 为您的留学生涯保驾护航 在代写量子场论Quantum field theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写量子场论Quantum field theory代写方面经验极为丰富，各种代写量子场论Quantum field theory相关的作业也就用不着说。

## 物理代写|量子场论代写Quantum field theory代考|Normalization and spin sums

To figure out what the normalization is that we have implicitly chosen, let us compute the inner product:

\begin{aligned} & \bar{u}s(p) u{s^{\prime}}(p)=u_s^{\dagger}(p) \gamma_0 u_{s^{\prime}}(p)=\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_s \ \sqrt{p \cdot \bar{\sigma}} \xi_s \end{array}\right)^{\dagger}\left(\begin{array}{ll} 0 & \mathbb{1} \ \mathbb{1} & 0 \end{array}\right)\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_{s^{\prime}} \ \sqrt{p \cdot \bar{\sigma}} \xi_{s^{\prime}} \end{array}\right) \ & =\left(\begin{array}{l} \xi_s \ \xi_s \end{array}\right)^{\dagger}\left(\begin{array}{ll} \sqrt{(p \cdot \sigma)(p \cdot \bar{\sigma})} & \ & \sqrt{(p \cdot \sigma)(p \cdot \bar{\sigma})} \end{array}\right)\left(\begin{array}{l} \xi_{s^{\prime}} \ \xi_{s^{\prime}} \end{array}\right) \ & =2 m \delta_{s s^{\prime}} \text {. } \ & \end{aligned}
Similarly, $\bar{v}s(p) v{s^{\prime}}(p)=-2 m \delta_{s s^{\prime}}$. This is the (conventional) normalization for the spinor inner product for massive Dirac spinors. It is also easy to check that $\bar{v}s(p) u{s^{\prime}}(p)=$ $\bar{u}s(p) v{s^{\prime}}(p)=0$.
We can also calculate
$$u_s^{\dagger}(p) u_{s^{\prime}}(p)=\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_s \ \sqrt{p \cdot \bar{\sigma}} \xi_s \end{array}\right)^{\dagger}\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_{s^{\prime}} \ \sqrt{p \cdot \bar{\sigma}} \xi_{s^{\prime}} \end{array}\right)=2 E \xi_s^{\dagger} \xi_{s^{\prime}}=2 E \delta_{s s^{\prime}},$$
and similarly, $v_s^{\dagger}(p) v_{s^{\prime}}(p)=2 E \delta_{s s^{\prime}}$. This is the conventional normalization for massless Dirac spinors. Another useful relation is that, if we define $\bar{p}^\mu=\left(E_p,-\vec{p}\right)$ as a momentum backwards to $p^\mu$, then $v_s^{\dagger}(p) u_{s^{\prime}}(\bar{p})=u_s^{\dagger}(p) v_{s^{\prime}}(\bar{p})=0$.
We can also compute the spinor outer product:
$$\sum_{s=1}^2 u_s(p) \bar{u}s(p)=\not p+m,$$ where the sum is over the spins. Both sides of this equation are matrices. It may help to think of this equation as $\sum_s|s\rangle\langle s|$. For the antiparticles, $$\sum{s=1}^2 v_s(p) \bar{v}_s(p)=\not p-m$$

## 物理代写|量子场论代写Quantum field theory代考|Majorana spinors

Recall from Section 10.6 that if we allow fermions to be anticommuting Grassmann numbers (these “numbers” will be discussed more formally in Section 14.6) then we can write down a Lagrangian for a single Weyl spinor with a mass term:
$$\mathcal{L}=i \psi_L^{\dagger} \sigma_\mu \partial_\mu \psi_L+i \frac{m}{2}\left(\psi_L^{\dagger} \sigma_2 \psi_L^{\star}-\psi_L^T \sigma_2 \psi_L\right)$$
The mass terms in this Lagrangian are called Majorana masses, and the Lagrangian is said to describe Majorana fermions. Majorana fermions transform under the same representations of the Lorentz group as Weyl fermions. The distinction comes in the quantum theory in which Majorana fermions are their own antiparticles. We will make this more precise through the notion of charge conjugation defined below.

It is sometimes useful to use the Dirac algebra to represent Majorana fermions, like we use it to describe Weyl fermions with the $P_{R / L}=\frac{1}{2}\left(1 \pm \gamma_5\right)$ projection operators. Majorana fermions can be put in four-component Dirac spinors as
$$\psi=\left(\begin{array}{c} \psi_L \ i \sigma_2 \psi_L^{\star} \end{array}\right) .$$
This transforms like a Dirac spinor because $\sigma_2 \psi_L^{\star}$ transforms like $\psi_R$. Then the Majorana mass can be written as
$$\frac{m}{2} \bar{\psi} \psi=i \frac{m}{2}\left(\psi_L^{\dagger} \sigma_2 \psi_L^{\star}-\psi_L^T \sigma_2 \psi_L\right),$$
which agrees with Eq. (11.33).
Note that (in the Weyl basis), using $\sigma_2^2=\mathbb{1}$,
\begin{aligned} -i \gamma_2 \psi^{\star} & =-i\left(\begin{array}{cc} 0 & \sigma_2 \ -\sigma_2 & 0 \end{array}\right)\left(\begin{array}{c} \psi_L \ i \sigma_2 \psi_L^{\star} \end{array}\right)^{\star}=\left(\begin{array}{c} (-i)(-i) \sigma_2 \sigma_2^{\star} \psi_L \ (-i)(-1) \sigma_2 \psi_L^{\star} \end{array}\right)=\left(\begin{array}{c} \psi_L \ i \sigma_2 \psi_L^{\star} \end{array}\right) \ & =\psi \end{aligned}
Let us then define the operation of charge conjugation $C$ by
$$C: \quad \psi \rightarrow-i \gamma_2 \psi^{\star} \equiv \psi_c,$$
where $\psi_c \equiv-i \gamma_2 \psi^{\star}$ means the charge conjugate of the fermion $\psi$. Thus, a Majorana fermion is its own charge conjugate.

## 物理代写|量子场论代写Quantum field theory代考|Normalization and spin sums

\begin{aligned} & \bar{u}s(p) u{s^{\prime}}(p)=u_s^{\dagger}(p) \gamma_0 u_{s^{\prime}}(p)=\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_s \ \sqrt{p \cdot \bar{\sigma}} \xi_s \end{array}\right)^{\dagger}\left(\begin{array}{ll} 0 & \mathbb{1} \ \mathbb{1} & 0 \end{array}\right)\left(\begin{array}{c} \sqrt{p \cdot \sigma} \xi_{s^{\prime}} \ \sqrt{p \cdot \bar{\sigma}} \xi_{s^{\prime}} \end{array}\right) \ & =\left(\begin{array}{l} \xi_s \ \xi_s \end{array}\right)^{\dagger}\left(\begin{array}{ll} \sqrt{(p \cdot \sigma)(p \cdot \bar{\sigma})} & \ & \sqrt{(p \cdot \sigma)(p \cdot \bar{\sigma})} \end{array}\right)\left(\begin{array}{l} \xi_{s^{\prime}} \ \xi_{s^{\prime}} \end{array}\right) \ & =2 m \delta_{s s^{\prime}} \text {. } \ & \end{aligned}

## 物理代写|量子场论代写Quantum field theory代考|Momentum-space Feynman rules

$$\mathcal{T}1=\overbrace{}^{x_1} \overbrace{}^{x_2} \bullet=-\frac{g^2}{2} \int d^4 x \int d^4 y D{1 x} D_{x y}^2 D_{y 2} .$$

$$D_{x y}=\int \frac{d^4 p}{(2 \pi)^4} \frac{i}{p^2+i \varepsilon} e^{i p(x-y)}$$

\begin{aligned} \mathcal{T}_1= & -\frac{g^2}{2} \int d^4 x \int d^4 y \int \frac{d^4 p_1}{(2 \pi)^4} \int \frac{d^4 p_2}{(2 \pi)^4} \int \frac{d^4 p_3}{(2 \pi)^4} \int \frac{d^4 p_4}{(2 \pi)^4} \ & \times e^{i p_1\left(x_1-x\right)} e^{i p_2\left(y-x_2\right)} e^{i p_3(x-y)} e^{i p_4(x-y)} \frac{i}{p_1^2+i \varepsilon} \frac{i}{p_2^2+i \varepsilon} \frac{i}{p_3^2+i \varepsilon} \frac{i}{p_4^2+i \varepsilon} \end{aligned}

\begin{aligned} & \mathcal{T}_1=-\frac{\lambda^2}{2} \int \frac{d^4 k}{(2 \pi)^4} \int \frac{d^4 p_1}{(2 \pi)^4} \int \frac{d^4 p_2}{(2 \pi)^4} e^{i p_1 x_1} e^{-i p_2 x_2} \ & \times \frac{i}{p_1^2+i \varepsilon} \frac{i}{p_2^2+i \varepsilon} \frac{i}{\left(p_1-k\right)^2+i \varepsilon} \frac{i}{k^2+i \varepsilon}(2 \pi)^4 \delta^4\left(p_1-p_2\right) \end{aligned}

$$\langle f|S| i\rangle=\left[-i \int d^4 x_1 e^{-i p_i x_1}\left(p_i^2\right)\right]\left[-i \int d^4 x_2 e^{i p_f x_2}\left(p_f^2\right)\right]\left\langle\Omega\left|T\left{\phi\left(x_1\right) \phi\left(x_2\right)\right}\right| \Omega\right\rangle,$$

$$\langle f|S| i\rangle=-\int d^4 x_1 e^{-i p_i x_1}\left(p_i\right)^2 \int d^4 x_2 e^{i p_f x_2}\left(p_f^2\right) \mathcal{T}_1+\cdots$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。